Dùng liên hợp.

pt <=> \(\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(1+\sqrt{3}\right)\)

\(-3\left(x-1\right)\left(x-\sqrt{3}\right)\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}\right)\)

\(+2\left(x-1\right)\left(x-\sqrt{2}\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}+\sqrt{2}\right)=3x-1\)

<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left[\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)-\left(x-1\right)\left(\sqrt{2}+\sqrt{3}\right)\right]\)

\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left[\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)-\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)\right]\)

\(=3x-1\)

<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(x+\sqrt{3}\right)\left(1-\sqrt{2}\right)\)

\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left(x+1\right)\left(\sqrt{2}-\sqrt{3}\right)=3x-1\)

<=> \(3-x^2-2\left(1-x^2\right)=3x-1\)

<=> \(x^2-3x+2=0\) phương trình bậc 2.

Em làm tiếp nhé!

Bài 1:

a/ \(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)

\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}\)

\(=\frac{10}{2}=5\)

b/ \(=\left(2-\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)

\(=\left(2-\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)

\(=\left(2-\sqrt{3}\right)\sqrt{2+4\sqrt{6}}\)

Bạn coi lại đề, tới đây ko rút gọn được nữa nên chắc bạn ghi đề nhầm ở chỗ nào đó

c/ \(=\frac{5\left(\sqrt{3}+\sqrt{2}\right)\left(5-\sqrt{24}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-\sqrt{24}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\)

\(=\left(5+2\sqrt{6}\right)\left(5-\sqrt{24}\right)=\left(5+\sqrt{24}\right)\left(5-\sqrt{24}\right)=1\)

d/ Nhân cả tử và mẫu của từng phân số với liên hợp của mẫu, mẫu số sẽ thành 1 hết:

\(=\frac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\frac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-1\)

\(=\sqrt{25}-1=5-1=4\)

Câu 2:

a/ ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}=2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=2\)

TH1: \(\sqrt{x-1}-1\ge0\Rightarrow x\ge2\) pt trở thành:

\(\sqrt{x-1}-1=2\Leftrightarrow\sqrt{x-1}=3\)

\(\Rightarrow x-1=9\Rightarrow x=10\) (nhận)

TH2: \(\sqrt{x-1}-1< 0\Rightarrow1\le x< 2\) pt trở thành:

\(1-\sqrt{x-1}=2\Rightarrow\sqrt{x-1}=-1< 0\) (vô nghiệm)

b/

\(\sqrt{x^2-x}+\sqrt{x^2+x-2}=0\)

Do \(\left\{{}\begin{matrix}\sqrt{x^2-x}\ge0\\\sqrt{x^2+x-2}\ge0\end{matrix}\right.\) nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\sqrt{x^2-x}=0\\\sqrt{x^2+x-2}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x^2+x-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x=1\)

\(A=\left(\frac{\sqrt{3}}{x^2+x\sqrt{x}+3}+\frac{3}{x^3-\sqrt{27}}\right)\left(\frac{x}{\sqrt{3}}+\frac{\sqrt{3}}{x}+1\right)\)

\(\Leftrightarrow A=\left[\frac{\sqrt{3}\left(x-\sqrt{3}\right)}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}+\frac{3}{\left(x-\sqrt{3}\right)\left(x+x\sqrt{3}+3\right)}\right]\left(\frac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\right)\)

\(\Leftrightarrow A=\frac{x\sqrt{3}-3+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}.\frac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)

\(\Leftrightarrow A=\frac{1}{x-\sqrt{3}}\)

Bài làm

Rút gọn

\(\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{1-x}\right)\cdot\frac{x-\sqrt{x}}{2\sqrt{x}+1}\)

\(=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)

\(=\left(\frac{\sqrt{x}+1}{(\sqrt{x}-1)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)

\(=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)

\(=\frac{\sqrt{x}}{\sqrt{x}+1}\)

Tính:

\(\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{21}{\sqrt{3}}\)

\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{7\sqrt{3}\cdot\sqrt{3}}{\sqrt{3}}\)

\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+7\sqrt{3}\)

\(=\frac{\left(3-\sqrt{3}\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)

\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{3+2\sqrt{3}}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)

\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}+3+2\sqrt{3}}{3-4}+7\sqrt{3}\)

\(=\frac{7\sqrt{3}-6}{-1}+7\sqrt{3}\)

\(=6-7\sqrt{3}+7\sqrt{3}\)

\(=6\)

Bài làm

\(\sqrt{42-10\sqrt{17}}+\sqrt{\left(\sqrt{17}-\sqrt{16}\right)^2}\)

\(=\sqrt{42-10\sqrt{17}}+\left|\sqrt{17}-\sqrt{16}\right|\)

\(=\sqrt{25-10\sqrt{17}+17}+\sqrt{17}-\sqrt{16}\)

\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{17}-\sqrt{16}\)

\(=\left|5-\sqrt{17}\right|+\sqrt{17}-\sqrt{16}\)

\(=5-\sqrt{17}+\sqrt{17}-\sqrt{16}\)

\(=5-4\)

\(=1\)

Bài 1:

a) Ta có: \(\sqrt{243}-\frac{1}{2}\sqrt{12}-2\sqrt{75}+\sqrt{27}\)

\(=\sqrt{3}\cdot9-\frac{1}{2}\cdot\sqrt{3}\cdot2-2\cdot\sqrt{3}\cdot5+\sqrt{3}\cdot3\)

\(=\sqrt{3}\left(9-1-10+3\right)\)

\(=\sqrt{3}\cdot1=\sqrt{3}\)

b) Ta có: \(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{5}{1+\sqrt{6}}-6\sqrt{\frac{1}{6}}\)

\(=\frac{\left(2\sqrt{3}-3\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(\sqrt{3}+\sqrt{2}\right)}+\frac{5\cdot\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\sqrt{36\cdot\frac{1}{6}}\)

\(=-\sqrt{6}+\frac{5\left(\sqrt{6}-1\right)}{5}-\sqrt{6}\)

\(=-2\sqrt{6}+\sqrt{6}-1\)

\(=-\sqrt{6}-1\)

Bài 2: Rút gọn

Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

2:

a: Sửa đề: \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

\(A=\dfrac{a^2+3}{\sqrt{a^2+2}}=\dfrac{a^2+2+1}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\)

=>\(A>=2\cdot\sqrt{\sqrt{a^2+2}\cdot\dfrac{1}{\sqrt{a^2+2}}}=2\)

A=2 thì a^2+2=1

=>a^2=-1(loại)

=>A>2 với mọi a

b: \(\Leftrightarrow\sqrt{a}+\sqrt{b}< =\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}\)

=>\(a\sqrt{a}+b\sqrt{b}>=a\sqrt{b}+b\sqrt{a}\)

=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)>=0\)

=>(căn a+căn b)(a-2*căn ab+b)>=0

=>(căn a+căn b)(căn a-căn b)^2>=0(luôn đúng)

1

ĐK: `x>1`

PT trở thành:

\(\sqrt{\dfrac{2x-3}{x-1}}=2\\ \Leftrightarrow\dfrac{2x-3}{x-1}=2^2=4\\ \Leftrightarrow4x-4-2x+3=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\left(KTM\right)\)

Vậy PT vô nghiệm.

b

ĐK: \(x\ge2\)

Đặt \(t=\sqrt{x-2}\) (\(t\ge0\))

=> \(x=t^2+2\)

PT trở thành: \(t^2+2-5t+2=0\)

\(\Leftrightarrow t^2-5t+4=0\)

nhẩm nghiệm: `a+b+c=0` (`1+(-5)+4=0`)

\(\Rightarrow\left\{{}\begin{matrix}t=1\left(nhận\right)\\t=4\left(nhận\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{x-2}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\x=18\left(TM\right)\end{matrix}\right.\)